Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n...












2












$begingroup$



Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n to int f$?




I know it is an application of DCT but I really don't know to come up with a bound that would satisfy the hypothesis of DCT. Something like $frac{1}{x^2}$ should work but $frac{1}{x} > frac{1}{x^2}$.






real-analysis limits measure-theory convergence lebesgue-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
    $endgroup$
    – mechanodroid
    Dec 14 '18 at 22:13










  • $begingroup$
    $int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
    $endgroup$
    – Kolja
    Dec 14 '18 at 22:13












  • $begingroup$
    what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
    $endgroup$
    – Infinity
    Dec 14 '18 at 22:20












  • $begingroup$
    We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
    $endgroup$
    – mechanodroid
    Dec 14 '18 at 22:23










  • $begingroup$
    Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
    $endgroup$
    – GEdgar
    Dec 14 '18 at 22:41
















2












$begingroup$



Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n to int f$?




I know it is an application of DCT but I really don't know to come up with a bound that would satisfy the hypothesis of DCT. Something like $frac{1}{x^2}$ should work but $frac{1}{x} > frac{1}{x^2}$.






real-analysis limits measure-theory convergence lebesgue-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
    $endgroup$
    – mechanodroid
    Dec 14 '18 at 22:13










  • $begingroup$
    $int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
    $endgroup$
    – Kolja
    Dec 14 '18 at 22:13












  • $begingroup$
    what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
    $endgroup$
    – Infinity
    Dec 14 '18 at 22:20












  • $begingroup$
    We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
    $endgroup$
    – mechanodroid
    Dec 14 '18 at 22:23










  • $begingroup$
    Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
    $endgroup$
    – GEdgar
    Dec 14 '18 at 22:41














2












2








2





$begingroup$



Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n to int f$?




I know it is an application of DCT but I really don't know to come up with a bound that would satisfy the hypothesis of DCT. Something like $frac{1}{x^2}$ should work but $frac{1}{x} > frac{1}{x^2}$.






real-analysis limits measure-theory convergence lebesgue-integral







share|cite|improve this question











$endgroup$





Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n to int f$?




I know it is an application of DCT but I really don't know to come up with a bound that would satisfy the hypothesis of DCT. Something like $frac{1}{x^2}$ should work but $frac{1}{x} > frac{1}{x^2}$.






real-analysis limits measure-theory convergence lebesgue-integral




real-analysis limits measure-theory convergence lebesgue-integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 22:20









Batominovski

33.1k33293




33.1k33293










asked Dec 14 '18 at 22:09









InfinityInfinity

346112




346112












  • $begingroup$
    As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
    $endgroup$
    – mechanodroid
    Dec 14 '18 at 22:13










  • $begingroup$
    $int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
    $endgroup$
    – Kolja
    Dec 14 '18 at 22:13












  • $begingroup$
    what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
    $endgroup$
    – Infinity
    Dec 14 '18 at 22:20












  • $begingroup$
    We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
    $endgroup$
    – mechanodroid
    Dec 14 '18 at 22:23










  • $begingroup$
    Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
    $endgroup$
    – GEdgar
    Dec 14 '18 at 22:41


















  • $begingroup$
    As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
    $endgroup$
    – mechanodroid
    Dec 14 '18 at 22:13










  • $begingroup$
    $int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
    $endgroup$
    – Kolja
    Dec 14 '18 at 22:13












  • $begingroup$
    what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
    $endgroup$
    – Infinity
    Dec 14 '18 at 22:20












  • $begingroup$
    We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
    $endgroup$
    – mechanodroid
    Dec 14 '18 at 22:23










  • $begingroup$
    Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
    $endgroup$
    – GEdgar
    Dec 14 '18 at 22:41
















$begingroup$
As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:13




$begingroup$
As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:13












$begingroup$
$int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
$endgroup$
– Kolja
Dec 14 '18 at 22:13






$begingroup$
$int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
$endgroup$
– Kolja
Dec 14 '18 at 22:13














$begingroup$
what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
$endgroup$
– Infinity
Dec 14 '18 at 22:20






$begingroup$
what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
$endgroup$
– Infinity
Dec 14 '18 at 22:20














$begingroup$
We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:23




$begingroup$
We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:23












$begingroup$
Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
$endgroup$
– GEdgar
Dec 14 '18 at 22:41




$begingroup$
Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
$endgroup$
– GEdgar
Dec 14 '18 at 22:41










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$begingroup$

as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.






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    $begingroup$

    as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.






        share|cite|improve this answer









        $endgroup$



        as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 22:36









        MalikMalik

        1289




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